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Find the intercepts

shamieh

Active member
Sep 13, 2013
539
Find the intercepts.

How do I find the y intercept?
\(\displaystyle
\frac{x^2}{\sqrt{x + 1}}\)

the x is easy I just plug in a 0.

for the y intercept what are the rules...It's like if the exponent is bigger than the bottom exponent then = undefined?? and if its the same you look at the coefficient and if it's smaller exponent/bigger exponent= 0?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
You have it backwards actually. To find the y-intercept you plug in 0 for x and solve. To find the x-intercept you plug in 0 for y and solve.

So you'll get two equations:

(1) \(\displaystyle 0 = \frac{x^2}{\sqrt{x + 1}}\)

(2) \(\displaystyle y=\frac{0^2}{0+1}\)

Which one is which?
 

shamieh

Active member
Sep 13, 2013
539
okay so y = 0..

so how do I find the x... I say x^2 = 0 and then x = 0 ? Oh I think I see now!

- - - Updated - - -

Jameson, since this is relating to the same problem...What about Vertical Asymptote and Horizontal Asymptotes?

for Vertical asymptote I would just set the denominator to zero correct?

x + 1 = 0 which means x = - 1 for V.A.


then for the horizontal asymptote how do I find it? if the exponent in the numerator is larger than the exponent in the denominator it would mean that it is undefined therefore being no Horizontal Asymptote? Am I Correct in saying this?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Yep, exactly! When we divide by zero we have a vertical asymptote so we set the denominator equal to 0. The exponent in the numerator is larger than the one in the denominator so we conclude that it doesn't stabilize to some horizontal value. (Yes).

Here is what it looks like by the way and you should see you're correct on all accounts:

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